Differentiation Question 85
Question: If $ u={\log _{e}}(x^{2}+y^{2})+{{\tan }^{-1}}( \frac{y}{x} ) $ , then $ \frac{{{\partial }^{2}}u}{\partial x^{2}}+\frac{{{\partial }^{2}}u}{\partial y^{2}}= $
[EAMCET 2000]
Options:
A) 0
B) 2u
C) 1/u
D) u
Show Answer
Answer:
Correct Answer: A
Solution:
$ u={\log _{e}}(x^{2}+y^{2})+{{\tan }^{-1}}( \frac{y}{x} ) $
$ \frac{\partial u}{\partial x}=\frac{2x}{x^{2}+y^{2}}+\frac{1}{1+\frac{y^{2}}{x^{2}}}.( -\frac{y}{x^{2}} ) $
$ =\frac{2x-y}{x^{2}+y^{2}} $
$ \frac{{{\partial }^{2}}u}{\partial x^{2}}=\frac{(x^{2}+y^{2}).2-(2x-y)2x}{{{(x^{2}+y^{2})}^{2}}} $
$ =\frac{2y^{2}-2x^{2}+2xy}{{{(x^{2}+y^{2})}^{2}}} $
$ \frac{\partial u}{\partial y}=\frac{2y}{x^{2}+y^{2}}+\frac{1}{1+\frac{y^{2}}{x^{2}}}.\frac{1}{x}=\frac{2y+x}{x^{2}+y^{2}} $
$ \frac{{{\partial }^{2}}u}{\partial y^{2}}=\frac{(x^{2}+y^{2}).2-(2y+x)2y}{{{(x^{2}+y^{2})}^{2}}} $ = $ \frac{2x^{2}-2y^{2}-2xy}{{{(x^{2}+y^{2})}^{2}}} $
$ \therefore $ $ \frac{{{\partial }^{2}}u}{\partial x^{2}}+\frac{{{\partial }^{2}}u}{\partial y^{2}}=0 $ .
BETA
